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  • A non - uniform bar of weight W is suspended at rest by two strings of negligible weight as shown.

Q7.8     A non-uniform bar of weight W is suspended at rest by two strings of negligible
             weight as shown in Fig.7.39. The angles made by the strings with the vertical are
             36.9^{0}and 53.1^{0}  respectively. The bar is 2m  long. Calculate the distance d  of the
           centre of gravity of the bar from its left end.

Answers (1)

best_answer

The FBD of the given bar is shown below : 

                                      Rotational motion, 20121

Since the bar is in equilibrium, we can write :

                                               T_1 \sin 36.9 ^{\circ}\ =\ T_2 \sin 53.1 ^{\circ}

or                                                            \frac{T_1}{T_2}\ =\ \frac{0.800}{0.600}\ =\ \frac{4}{3}

or                                                            T_1\ =\ \frac{4}{3}T_2                                    .....................................................(i)

For the rotational equilibrium :

                                                   T_1 \cos 36.9 ^{\circ}\times d\ =\ T_2 \cos 53.1 ^{\circ}\times \left ( 2-d \right )                       (Use equation (i) to solve this equation)

or                                                                             d\ =\ \frac{1.2}{1.67}\ =\ 0.72\ m

Thus the center of gravity is at 0.72 m from the left.

Posted by

Devendra Khairwa

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